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Andru [333]
3 years ago
12

tech a says that a tire with more wear on the center of the tread is caused by under inflation of the size tech b says featherin

g of the tire tread is most commonly a result of a excessive toe in or toe out who is correct answer
Engineering
1 answer:
Natali5045456 [20]3 years ago
4 0

Answer: Technician B only is correct.

Explanation:

Tire tread wear on the edges of a tire will typically indicate inflation pressures are lower than specified.

Under inflation occurs when there is more wear on the edges of the tire. A tire is said to be under-inflated whe the contact patch grows and outside edges of the patch takes on the load.

Tire feathering or scuffing is the indicator of excessive positive or negative toe angle that can be detected by stroking your fingertips over the edge of each tread block. A feather edge on the inside of the tread bar shows excess toe-in, whereas a feather edge on the outside of the tread bar shows toe-out. Changes in camber and caster angle affects the toe angle. Changes in suspension height can also affect the toe angle geometry.

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A bridge a mass of 800 kg and is able to support up to 4 560 kg. What is its structural efficiency?
fiasKO [112]

Answer:4560÷800=5.4

Explanation:

3 0
2 years ago
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What are the mechanical properties of a geotextile that are of most importance when using it as a separator in an unpaved road s
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3 years ago
A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi
Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

F_{e} = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

                                      = 7.5 x 1000 W

                                      = 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 =  F₁ - F₂

750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

∴F₂  = 750 N

Now F₁ = 2 F₂

        F₁ = 2 x F₂

        F₁ = 2 x 750

        F₁ = 1500 N   ,   this is the maximum force.

Therefore we know,

F_{max} = 3 x F_{e}

where F_{e} is centrifugal force

 F_{e} = F_{max} / 3

                          = 1500 / 3

                         = 500 N

8 0
3 years ago
What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP
ella [17]

Answer:

The modulus of resilience is 166.67 MPa

Explanation:

Modulus of resilience is given by yield strength ÷ strain

Yield strength = 500 MPa

Strain = 0.003

Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa

3 0
3 years ago
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A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant spee
Ludmilka [50]

Answer:

S = 0.5 km

velocity of motorist = 42.857 km/h

Explanation:

given data

speed  = 70 km/h

accelerates uniformly = 90 km/h

time = 8 s

overtakes motorist =  42 s

solution

we know  initial velocity u1 of police = 0

final velocity u2 = 90 km/h = 25 mps

we apply here equation of motion

u2 = u1 + at  

so acceleration a will be

a = \frac{25-0}{8}

a = 3.125  m/s²

so

distance will be

S1 = 0.5 × a × t²

S1 = 100 m = 0.1 km

and

S2 = u2 ×  t

S2 = 25  × 16

S2 = 400 m = 0.4 km  

so total distance travel by police

S = S1 + S2

S = 0.1 + 0.4

S = 0.5 km

and

when motorist travel with  uniform velocity

than total time = 42 s

so velocity of motorist will be

velocity of motorist = \frac{S}{t}

velocity of motorist =  \frac{500}{42}  

velocity of motorist = 42.857 km/h

3 0
3 years ago
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